Let $E/\mathbf{Q}$ be an elliptic curve of rank $>0$. It is easy to see that there is a positive-density set of primes $p$ such that the reduction map $\mathrm{red}_p : E(\mathbf{Q}) \rightarrow \widetilde{E}(\mathbf{F}_p)$ is not surjective. Namely, for an integer $n>1$, take any rational prime $p$ that splits completely in the field $K_n = \mathbf{Q}([n]^{-1}(E(\mathbf{Q})))$, i.e. the field that results from adjoining to $\mathbf{Q}$ the coordinates of all preimages of points in $E(\mathbf{Q})$ under multiplication by $n$. Note that the $K_n$ are finite field extensions of $\mathbf{Q}$ (this is equivalent to the weak Mordell-Weil theorem). To show that these $p$ work: take $P \in E(\mathbf{Q})$ and $Q \in E(K_n)$ with $nQ=P$, then $\mathrm{red}_p(P) = n ( \mathrm{red}_p(Q))$, with $\mathrm{red}_p(Q)$ lying in $\widetilde{E}(\mathbf{F}_p)$ by the assumption on $p$, so $\mathrm{red}_p(E(\mathbf{Q}))$ lies in $n \widetilde{E}(\mathbf{F}_p)$, which is an index-$n^2$ subgroup of $\widetilde{E}(\mathbf{F}_p)$.

More generally, for any isogeny $\phi : E' \rightarrow E$ of elliptic curves over $\mathbf{Q}$ with non-trivial kernel, take any prime $p$ that splits completely in the finite field extension $\mathbf{Q}(\phi^{-1}(E(\mathbf{Q})))$ of $\mathbf{Q}$.

My questions are in the opposite direction:

Do there exist infinitely many $p$ such that $\mathrm{red}_p$ is surjective?

Do there exist arbitrarily large sets of primes $\{ p_1, p_2, \ldots, p_m \}$ such that the combined reduction map $E(\mathbf{Q}) \rightarrow \prod_{i=1}^m \widetilde{E}(\mathbf{F}_{p_i})$ is surjective?

For a prime $p$ such that $\mathrm{red}_p$ is not surjective, is the failure of surjectivity explained by some isogeny $\phi$, by the argument sketched in the first paragraph?

Edit. The answer to 1. and 2. is obviously "no" in general by Maarten's answer. By the Gupta-Murty paper mentioned by Felipe, the answer to 1. becomes "yes" once the rank of $E$ is sufficiently large. As for 2., I would like to ask: is there even a single elliptic curve $E$ over $\mathbf{Q}$ for which question 2. has a positive answer?

2 Answers
2

Part 1 is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.

Let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$.
Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.

A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at https://sage.mderickx.nl/home/pub/9

Update on part 1:

I extended the search to rank > 1 curves and also found multiple examples of an isogeny between rank 2 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$. An example is where $E'$ is the elliptic curve defined by $y^2 + xy + y = x^3 + x^2 - 71x - 196$ and the kernel of $\phi$ is generated by $(9 : -5 : 1)$. The Cremona label of $E'$ is '3315b2'.
I did not find any examples of rank 3 after searching trough all elliptic curves of conductor < 100000.

ps. Note that my counter examples to part one are not counter examples to Lang-Trotter. The reason is that the Lang-Trotter conjecture is a conjecture about the density of the of the primes such that the reduction is surjective. In my examples both the conjectured density and the actual density are both 0.

Part 3:
Let $E$ be a non CM rank 1 elliptic curve that is the only one in it's isogeny class. Then the only isogenies to $E$ are the multiplication by $n$ maps. For concreteness I let $E$ be the curve among all curves with these properties of smallest conductor. This curve $E$ is given by $y^2 + y = x^3 - x$ and $E(\mathbb Q)= \mathbb Z$ is generated by $P=(0 : -1 : 1)$. Now let $p=23$ then $\\#E(\mathbb F_p)=22$ but the order of $P$ after reduction is $11$ so that the index is $2$. This means that the obstruction cannot come from an isogeny because this would mean that it comes from some multiplication by $n$ map and hence that the index should not be squarefree.

In the article of Lang and Trotter where they state their conjecture they give a criterion in terms of $\mathbb Q(l^{-1}E(\mathbb Q))$ that is equivalent to $l$ being a divisor of the index. If you read that obstruction carefully you will realize that its really easy to cook up counter examples to part $3$, in particular using their criterion one can show that the set of $p$ such that reduction mod $p$ is a counter example has positive density for the above $E$.

I'm still continuing my search for more counter examples :). Looking at different sextic twists of y^2=x^3+1 I also found a CM rank 3 counter example to part 1. This counter example is interesting since the Gupta-Murty paper proves that Part 1 holds for CM curves of rank $\geq 6$. So this counter example shows that the Gupta-Murty result at least needs something like rank $\geq 4$ as a condition. In this counter example $E'$ is given by $y^2 =x^3 + 14683622976$ and $\phi$ given by dividing out the group of order $3$ generated by $(0 : 121176 : 1)$.
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Maarten DerickxNov 25 '12 at 16:22

So far, all your counterexamples seem to come from isogenies of degree $3$. Any reason to expect isogenies $\phi$ of degree $2$ not to satisfy $\mathbf{Q}(\phi^{-1}E(\mathbf{Q})) = \mathbf{Q}$?
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RenéNov 25 '12 at 21:46

No, in my large search trough the entire Cremona Database for counter exmamples I found counter examples with isogenies of prime degree 2,3,5 and 7. So it seems that there is not much of an obstruction comming from the kind of isogeny.
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Maarten DerickxNov 25 '12 at 21:57

Part 1. is known if the rank of $E(\mathbb{Q})$ is sufficiently large (Gupta, Murty, Primitive points on elliptic curves. Compositio Math. 58 (1986), 13–44, is the original result, which may have been improved). The case of rank one is the elliptic analogue of Artin's conjecture on primitive roots and is open. Part 2 seems harder. Part 3 seems obvious to me, just look at an isogeny multiplication by the index of the image of reduction in the group of points modulo $p$.

Hi Felipe, I don't mind getting all these answers, they've been very helpful :) I've checked out the reference to Gupta-Murty, which does indeed prove what you say. Rank $\ge 6$ is large though, so I'll check for improvements later. (Although if Maarten Derickx's calculation proves correct, I guess some assumption on the size of $E(\mathbf{Q})$ on top of rank $>0$ is necessary.) I don't understand your solution to 3., do you mean to start by picking an isogeny that maps to the reduced curve? If so, I fail to see how you can get anything from that unless the isogeny lifts to $\mathbf{Q}$.
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RenéNov 21 '12 at 0:50

1

I was thinking about the cyclic case. As in Maarten's answer, the non-cyclic situation is trickier. If $P$ is a generator and doesn't map to a generator, then either the reduction has a non-cyclic group (and the prime $p$ splits in the field adjoining the relevant torsion) or the reduction is cyclic of larger order and the image of $P$ is divisible by the index and you can detect that by how the prime $p$ behaves in the extension obtained by dividing $P$ by the index.
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Felipe VolochNov 21 '12 at 2:57